BEEM103 – Optimization Techniques for Economists Dieter Balkenborg
Homework Week 3 – SOLUTIONS
Departments of Economics
University of Exeter
Exercise 1 Show that the function
Q = K L with ; > 0 and + < 1 is concave.
Solution 1 It is assmed here that K and L are positive.
H =
"
Since 0 < < 1 we have
@
2
Q
@K 2
@
2
Q
@K@L
@
2
Q
@K 2
= (
@Q
@K
= K
#
@
2
Q
@L@K
@
2
Q
@L 2
=
1) K
1
(
L ,
K
@Q
= K L
@L
1) K
1 L
2
1
L
2
L < 0 . Moreover det H =
=
(
(
= [ (1
1
(
K 1 L
1) K L
1) ( 1)
+ 1)
2 2
2 2
K
2 2
L
2 2
K
)] K
2 2
L
2 2
> 0
2 2
L
2 2
1
2
Hence the function is concave.
Exercise 2 Use the least-square criterion to …nd the equation of the line that is closest to the points
(1 ; 1) , (2 ; 3) and (4 ; 3) : That is, …nd the line y = f ( x ) = ax + b which minimizes
F ( a; b ) = i =1
( y i f ( x i
))
2 where the summation is over the above three data points ( x i
; y i
) .
Solution 2 The …rst order-conditions are with nx = i =1 x i
, ny = i =1 y i
, nv = i =1 x
2 i
, nc = i =1 x i y i
@F
@b
@F
@a
= 2 i =1
( y i ax i b ) = 0 () ax + b = y
= 2 i =1
( y i ax i b ) ( x i
) = 0 () av
The solution is given by ax
2
+ bx = xy av + bx = c a v x
2
= c xy a = b = y ax = y c xy
( v x 2 ) x
Now put in the numbers to get a = 4
7 and b = 1 in our example.
+ bx = c

Exercise 3 The highway department is planning to build a picnic area for motorists along a major highway. It is to be rectangular with an area of 5 ; 000 square yards and is to be fenced o¤ on the three sides not adjacent to the highway. What is the least amount of fencing that will be needed to complete the job?
a) Identify this problem as a constraint optimization problem. What objective function f ( x; y ) is to be maximized / minimized subject to what constraint g ( x; y ) 0 ?
b) Write down the Lagrangian L ( x; y ) for this problem.
c) Find the solution to the three equations a)
@ L
@x
= 0 b)
@ L
@y
= 0 c) g ( x; y ) 0 .
Solution 3 a) Let x denote the length of the side of the rectangle adjacent to the highway and let y denote the length of the other side. Then we have to minimize the amount of fencing x + 2 y subject of the constraint that the area xy is at least 5000 ( xy 5000 ). Equivalently, we have to maximize x 2 y subject to the constraint xy 5000 0 .
b) The Lagrangian is
L ( x; y ) = x 2 y + ( xy 5000) c)
@ L
@x
@ L
= 1 + y = 0
=
@y
5000 = xy
2 + x = 0 or or
1 =
2 = y x
Division of the two equations on the right yields y 2
1
2
= y x
= y x or x = 2 y . Hence 5000 = xy = 2 y 2
= 2500 , y = 50 (since the solution 50 does not make sense). We obtain x = 100 , y = 50 and
,
= 1 = 50 0 .
Exercise 4 A consumer has the following utility when he consumes x units of apples and y units of oranges: u ( x; y ) = x
2
+ 4 x y
2
+ 16 y
Suppose the consumer has a budget of £ 3.20 to be spend on oranges and apples. Each apple and each orange costs £ 0.40. Use the method of Lagrange to …nd the optimal consumption bundle: a) Write down the budget constraint and the Lagrangian. Assume that only the budget constraint is binding.
Solution 4 a) With the given prices and budget the budget constraint is
0 : 4 x + 0 : 4 y = 3 : 2 .
The Lagrangian for the constrained utility maximization problem is hence
L ( x; y ) = u ( x; y ) + [3 : 2 0 : 4 x 0 : 4 y ] +
2 x +
3 y
= x
2
+ 4 x y
2
+ 16 y + [3 : 2 0 : 4 x 0 : 4 y ] since
2
=
3
= 0 by the complementarity conditionsb) In the optimum the two partial derivatives of the
Lagrangian must be zero:
@ L
@x
@ L
@x
=
=
2
2 x y
+ 4
+ 16
[0 :
[0
4] = 0
:
Division of the two equations on the right yields
4] = 0 or or
2 x
2 y
+ 4 = 0 : 4
+ 16 = 0 : 4
2 x + 4 0 : 4
=
2 y + 16 0 : 4
2 x 12 = 2 y
= 1 or or y = x
2
+ 6 x + 4 = 2 y + 16 or
2

Thus y = x + 6 must hold at the constrained utility maximum.
c) In addition, the constraint
0 : 4 x + 0 : 4 y = 3 : 2 y = 8 x or x + y =
3 : 2
0 : 4
= 8 must hold at the optimum. Overall y = x + 6 and y = 8 x must hold, so x + 6 = 8 x or 2 x = 2 or x = 1 or
Moreover, x = 1 implies y = 8 x = 7 .
Thus the Lagrangian approach suggests that it is optimal for the consumer to buy one apple and seven oranges.
(Notice that we obtain 0 : 4 = 16 2 y = 16 14 = 2 > 0 for the Lagrange multiplier.)
3